Question

Lithium nitride reacts with water to produce ammonia and lithium hydroxide according to the equation

Li₃N(s)+3H₂O(l)⟶NH₃(g)+3LiOH(aq)
Heavy water is water with the isotope deuterium in place of ordinary hydrogen, and its formula is D₂O.
The same reaction can be used to produce heavy ammonia, ND₃(g), according to the equation
Li₃N(s)+3D₂O(l)⟶ND₃(g)+3LiOD(aq)
Calculate how many grams of heavy water are required to produce 160.0 mg ND₃(g). The mass of deuterium, D, is 2.014 g/mol.

Answer 1

0.48 g D2O

Step-by-step explanation:

Using the balanced equation we can calculate the mol and grams of heavy water required to produce the 160 mg of ND3 as follows:

Li3N (s) + 3 D2O (l) -------------------- ND3 (g) + 3 LiOD (aq)

gD2O = ?

MW ND3 = 20.04 g/mol

160 mg ND3 = 0.160 g ND3 = 0.160g /20.04g /mol = 0.08 mol ND3

1 mol ND3 needs 3 mol D2O to be produced, then

0.08 mol ND3 x 3 mol D2O/1 mol ND3 = 0.024 mol D2O

Then multiplying this number of moles by the molecular weight of D2O ( 20.028 g D2O/mol ) we will get the answer:

0.024 mol D2O x 20.028 g D2O/ mol = 0.48 g D2O